Answer 2 & 16 Marks HMT.doc

8/9/2019 Answer 2 & 16 Marks HMT.doc

1/85

ME 2251 Heat and Mass Transfer

Unit-1 Conduction

Part-A

1. State Fourier’s Law of conduction. (

The rate of heat conduction is proportional to the area measured – normal to the

direction of heat flow and to the temperature gradient in that direction.

Qα - A dT

dx

dT - KA

dxQ =

where A – are in m2

dT

dx - Temperature gradient in K/m K – Thermal conductivity W/mK.

2. Define Thermal Conductiit!.

Thermal conductivity is defined as the aility of a sustance to conduct heat.

". #rite down the e$uation for conduction of heat throu%h a sla& or 'lane wall.

!eat transfer overall T Q R

∆= Where ∆ T " T# – T2

L R

KA= - Thermal resistance of sla

$ " Thic%ness of sla& K " Thermal conductivity of sla& A " Area

. #rite down the e$uation for conduction of heat throu%h a hollow c!linder.


∆= Where& ∆ T " T# – T2

2

1

r 1 in

2 R

LK r π

=

thermal resistance of sla

$ – $ength of cylinder& K – Thermal conductivity& r 2 – 'uter radius & r # – inner radius

). State *ewton’s law of coolin% or conection law.

!eat transfer y convection is given y (ewton)s law of cooling

Q " hA *Ts - T∞+

Where

A – Area e,posed to heat transfer in m2 & h - heat transfer coefficient in W/m2K

Ts – Temperature of the surface in K& T∞ - Temperature of the fluid in K.


2/85

+. #rite down the %eneral e$uation for one dimensional stead! state heat transfer in sla&

or 'lane wall with and without heat %eneration.

2 2 2

2 2 2

1

T T T T

t x y z

∂ ∂ ∂ ∂+ + =

∞ ∂∂ ∂ ∂

2 2 2

2 2 2

1

T T T q T

K t x y z α

∂ ∂ ∂ ∂+ + + =

∂∂ ∂ ∂

,. Define oerall heat transfer co-efficient.

The overall heat transfer y comined modes is usually e,pressed in terms of an overall

conductance or overall heat transfer co-efficient ).

!eat transfer Q " A ∆T.

. #rite down the e$uation for heat transfer throu%h com'osite 'i'es or c!linder.


∆=

&Where & ∆ T " Ta – T&

2 12

1 2

1 1 2 3

1 1 1 .

2a b

r r In In L

r r R

L h r K K h r π

= + + +

. #hat is critical radius of insulation (or/ critical thic0ness

ritical radius " r c ritical thic%ness " r c – r #

Addition of insulating material on a surface does not reduce the amount of heat transfer

rate always. 0n fact under certain circumstances it actually increases the heat loss up to certain

thic%ness of insulation. The radius of insulation for which the heat transfer is ma,imum is called

critical radius of insulation& and the corresponding thic%ness is called critical thic%ness.

1. Define fins (or/ e3tended surfaces.

0t is possile to increase the heat transfer rate y increasing the surface of heat transfer.

The surfaces used for increasing heat transfer are called e,tended surfaces or sometimes

%nown as fins.

11. State the a''lications of fins.

The main applications of fins are

#. ooling of electronic components2. ooling of motor cycle engines.1. ooling of transformers. ooling of small capacity compressors

12. Define Fin efficienc!.

The efficiency of a fin is defined as the ratio of actual heat transfer y the fin to the

ma,imum possile heat transferred y the fin.

max

fin

fin

Q

Qη =


3/85

1". Define Fin effectieness.

3in effectiveness is the ratio of heat transfer with fin to that without fin

3in effectiveness "with fin

without fin

Q

Q

4art -5

1. A wall is constructed of seeral la!ers. The first la!er consists of masonr! &ric0 2

cm. thic0 of thermal conductiit! .++ #4m56 the second la!er consists of " cm thic0

mortar of thermal conductiit! .+ #4m56 the third la!er consists of cm thic0 lime stone

of thermal conductiit! .) #4m5 and the outer la!er consists of 1.2 cm thic0 'laster of

thermal conductiit! .+ #4m5. The heat transfer coefficient on the interior and e3terior

of the wall are ).+ #4m

2

5 and 11 #4m

2

5 res'ectiel!. 7nterior room tem'erature is 22°Cand outside air tem'erature is -)°C.

Calculate

a/ 8erall heat transfer coefficient&/ 8erall thermal resistancec/ The rate of heat transfer d/ The tem'erature at the 9unction &etween the mortar and the limestone.

:ien Data

Thic%ness of masonry $# " 26cm " 6.26 m

Thermal conductivity K# " 6.77 W/mK

Thic%ness of mortar $2 " 1cm " 6.61 m

Thermal conductivity of mortar K2 " 6.7 W/mK

Thic%ness of limestone $1 " 8 cm " 6.68 m

Thermal conductivity K1 " 6.98 W/mK

Thic%ness of 4laster $ " #.2 cm " 6.6#2 m

Thermal conductivity K " 6.7 W/mK

0nterior heat transfer coefficient ha " 9.7 W/m2K

:,terior heat transfer co-efficient h " ## W/m2K

0nterior room temperature Ta " 22° ; 2


4/85

Solution;

!eat flow through composite wall is given y

overall T Q

R

∆= >3rom e?uation *#1+@ *or+ >!T Bata oo% page (o. 1@

Where& ∆ T " Ta – T

31 2 4

1 2 3 4

31 2 4

1 2 3 4

1 1

1 1

295 2!"

1 #.2# #.#3 #.#! #.#12 1

5. #. #. #.5! #. 11

a b

a b

a b

L L L L R

h A K A K A K A K A h A

T T Q

L L L L

h A K A K A K A K A h A

Q A

= + + + + +

−⇒ =

+ + + + +

−⇒ =

+ + + + +

2!eat transfer per unit area Q/A " 1.97 W/m

We %now& !eat transfer Q " A *Ta – T+ >3rom e?uation *#+@

Where – overall heat transfer co-efficient

$ %

34.5

295 2!

a b

QU

A T T

U

K

⇒ =× −

⇒ =−

2'verall heat transfer co - efficient " #.28 W/m

We %now

'verall Thermal resistance *C+

31 2 4

1 2 3 4

1 1

a b

L L L L R

h A K A K A K A K A h A= + + + + +

3or unit Area

31 2 4

1 2 3 4

1 1

1 #.2# #.#3 #.#! #.#12 1 &

5 #. #. #.5! #. 11

#.'! "

a b

L L L L R

h K K K K h

R K W

= + + + + +

+ + + + +

=

7nterface tem'erature &etween mortar and the limestone T"

0nterface temperatures relation


5/85

1 2 3 3 4 4 5 51 2

1 2 3 4

1

1

a

1

1

1

1 2

1

295-T 1 ( & )

1"

295"1"

29534.5

1"5.

2!!.!

a b

a b

a

a

a a

a

T T T T T T T T T T T T Q

R R R R R R

T T Q

R

h A h A

T Q Ah

T

T K

T T Q

R

− − − − −−⇒ = = = = = =

−⇒ =

=

−⇒ =

−⇒ =

⇒ =

−⇒ =

Q

2 1

1

1 1

1

2!!.! )

T LQ

L k A

K A

−= =

Q

2

1

1

2

2

2 3

2

3 2

2

2 2

2

3

2

2

3

3

2!!.!"

2!!.!34.5

#.2#

#.

2'!.3 K

T &

2'!.3 )

2'!.3"

2'!.334.5

#.#3

#.

2'.5 K

T Q A

L

K

T

T

T Q

R

T LQ

L K A

K A

T Q A

L

K

T

T

−⇒ =

−⇒ =

⇒ =

−⇒

−= =

−⇒ =

−⇒ =

⇒ =

Q

Temperature etween ortar and limestone *T1 is 2


6/85

:ien Data

Thic%ness of fire plate $# " 3rom !T data oo% page (o.#@

!eat flow overall T

Q R

∆= & Where

∆ T " Ta – T

31 2

1 2 3

a

31 2

1 2 3

1 1

T (&

1 1

a b

b

a b

L L L R

h A K A K A K A h A

T

L L L

h A K A K A K A h A

= + + + +

−⇒

+ + + +

>The term $1 is not given so neglect that term@


7/85

a

31 2

1 2 3

a

1 2

1 2

2

T (&

1 1

T ( &

1 1

923 3##"

1 #.#'5 #.##5 1

# 1.#35 53. !

" 29#'.'9 "

b

a b

b

a b

T

L L L

h A K A K A K A h A

T

L L

h A K A K A h A

Q A

Q A W m

−⇒

+ + + +

−⇒

+ + +

−=

+ + +

=

1The term $ is not given so neglect that term@

(ii/ 8utside surface tem'erature T"We %now that& 0nterface temperatures relation

1 2 3 31 2

1 2

3

*

3

......$ %

$ %

1 )

1

a b a b

a b

b

b

b

b

b

T T T T T T T T T T Q A

R R R R R

T T A Q

R

h A

T T Q

h A

− − − −−= = = = =

−⇒ =

=

−⇒ =

where

3

3

3

T ("A &

1

3##29#'.'9

1

!

3.4'3

b

b

T

h

T

T K

−⇒

−⇒ =

=

". A steel tu&e (5 > ".2+ #4m5/ of ). cm inner diameter and ,.+2 cm outer diameter is coered with 2.) cm la!er of insulation (5 > .2 #4m5/ the inside

surface of the tu&e receiers heat from a hot %as at the tem'erature of "1+ °C with

heat transfer co-efficient of 2 #4m25. #hile the outer surface e3'osed to the

am&ient air at "°C with heat transfer co-efficient of 1, #4m25. Calculate heat loss

for " m len%th of the tu&e.:ien

Dteel tue thermal conductivity K# " 1.27 W/mK


8/85

0nner diameter of steel d# " 9.68 cm " 6.6968 m0nner radius r # " 6.629 m'uter diameter of steel d2 "


9/85

!eat transfer #1

r

2

iT T

Q

Inr

KLπ

∞−=

>3rom e?uation (o.*1+@

onsidering h e the outside heat transfer co-efficient.

π

π

π π

∞

∞

−∴

+

=

−⇒ =

+

i

6

#

6

6 6

i

6

#

6

T TQ "

r 0n

r #

2 K$ A h

!ere A 2 r $

T TQ

r 0n

r #

2 K$ 2 r $h

To find the critical radius of insulation& differentiate Q with respect to r 6 ande?uate it to Eero.

i 2

6 6

6 6

# 6

i

2

6 6

6 c

# #6 *T T +

2 K$r 2 h$r dQ

dr r # #0n

2 K$ r 2 h$r

since *T T + 6

# # 62 K$r 2 h$r

K r r

h

π π

π π

π π

∞

∞

− − −

⇒ =

+

− ≠

⇒ − =

⇒ = =

9. A wire of + mm diameter with 2 mm thic0 insulation (5 > .11 #4m5/. 7f theconectie heat transfer co-efficient &etween the insulatin% surface and air is 2)#4m2L6 find the critical thic0ness of insulation. And also find the 'ercenta%e of chan%e in the heat transfer rate if the critical radius is used.

:ien Data

d#" 7 mmr # " 1 mm " 6.661 mr 2 " r # ; 2 " 1 ; 2 " 9 mm " 6.669 mK " 6.## W/mKh " 29 W/m

2K


10/85

Solution ;

#. ritical radius cK

r h

= >3rom e?uation (o.*2#+@

1

c

1c

6.##r . #6 m

29

r . #6 m

−

−

= = ×

= ×

ritical thic%ness " r c – r # 1

1

. #6 6.661

#. #6 m

−

−

= × −

= ×-1

critical thic%ness t " #. #6 *or+ #. mm×

2. !eat transfer through an insulated wire is given y

a

#

2

#

# 2

T T

Q r 0n

r # #

2 $ K h r π

−

=

+

[ ]

a

a

3rom !T data oo% 4age (o.19

2 $ *T T + "

6.6690n

#6.661

6.## 29 6.669

2 $ *T T +Q# "

#2.7

π

π

−

+ ×

−

!eat flow through an insulated wire when critical radius is used is given y

[ ]a 2 2 cc

#

# c

T TQ r r

r 0n

r # #2 $ K h r π

−= →

+


11/85

a

1

1

a 2

2 $ *T T + "

. #60n

6.661 #

6.## 29 . #6

2 $ *T T +Q "

#2.9


12/85

>3rom !T data oo% 4age (o.#@

(i/ Tem'erature at the end of the fin6 Put 3 > L

T - T cos h m >$-$@*A+

T T cos h *m$+T - T #

...*#+T T cos h *m$+

where

h4 m "

KA

4 " 4erimeter " 2 $ *Appro,+

" 2 6.696

4 " 6.# m

∞

∞

∞

∞

⇒ =

−⇒ =

−

××

A – Area " $ength × thic%ness " 6.696 × 6.66<

2 A 1.9 #6 m−= ×

h4 m "

KA⇒

#6 6.#

99 1.9 #6

m 27.=7

−

×=

× ×

=

, $

T - T #*#+

T T cos h *27.= 6.696+

T - T #

T T 2.69

T - 2=9 #

1=1 - 2=9 2.69

T - 2=9 "


13/85

[ ]

T - T cos hm >$-$/2@*A+

T T cos h *m$+

6.696cos h 27.= 6.696 -

T - T 2

T T cos h 27.= *6.696+

T- 2=9 #.21 1=1 - 2=9 2.6=

T - 2=9 6.7629

1=1 -2=9

T 19.6 K

∞

∞

∞

∞

⇒ =−

⇒ =

− ×

⇒ =

⇒ =

=

Temperature at the middle of the fin

, $ / 2T 19.6 K= =

(iii/ Total heat dissi'ated

>3rom !T data oo% 4age (o.#@

#/2

- #/ 2

Q " *h4KA+ *T T + tan h *m$+

>#6 6.# 99 1.9 #6 @ *1=1 2=9+

tan h *27.= 6.696+

Q " . W

∞⇒ −

⇒ × × × × × −× ×


14/85

>3rom !T data oo% 4age (o.2@Solution;

Thermal conductivity of the copper K " 187 W/mK3or sla&

haracteristic length c$

$

2

=

c

6.662 "

2

$ 6.66# m=

We %now&

5iot numer cih$

5K

=

#66 6.66#

187

×=

5i "2.9= #6 6.#−× <

5iot numer value is less than 6.#. Do this is lumped heat analysis type prolem.

3or lumped parameter system&

hAt

H

6

T Te

T T

ρ ρ

−×

× × ∞

∞

−=

−III.*#+

>3rom !T data oo% 4age (o.8@We %now&

haracteristics length $c "

H

A

c

ht

$

6

#66t

176 6.66# 8866

T-T*#+ e

T T

121 - 161 e7 1#4m25.B.. A'ril-26 216 226 harathi!ar ni. A'ril E harathi!ar ni. A'ril E


15/85

:ien

Dpecific heat ρ " 6.7 %G/%g K " 76 G/%g K

Thermal conductivity K " 19 W/mKBiameter of the sphere B " 9 cm " 6.69 mCadius of the sphere C " 6.629 m

0nitial temperature T6 " 96° ; 23rom !T data oo% 4age (o.#@Solution

Bensity of steel is 3rom !T data oo% 4age (o.8@We %now&

haracteristics length $c "H

A


16/85

c

1

ht

$

6

#6t

76 8.11 #6 .) 0@40% 56 h > 1) 5@4 hr m2 5. Determine (i/ Tem'erature of &all after 1

second and (ii/ Time for &all to cool to °C.

:ien

Biameter of the all B " #2 mm " 6.6#2 mCadius of the all C " 6.667m

0nitial temperature T6 " 866° ; 2 G/s " W@

×=

= Q

Bensity ρ "


17/85

We %now&

5iot numer cih$

5K

=

#.77< 6.662

97.=

×=

5i " #.7 × #6-1 J 6.#

5iot numer value is less than 6.#. Do this is lumped heat analysis type prolem.

3or lumped parameter system&

hAt

H

6

T Te

T T

ρ ρ

−×

× × ∞

∞

−=

−III.*#+

>3rom !T data oo% 4age (o.8@We %now&

haracteristics length $c " H A

c

ht

$

6

#.77<#6

96 6.662


18/85

Ta0e 5 for steel > 2.) #4m56 for steel > ." m2 4hr.

:ienThic%ness $ " 9 cm " 6.69 m

0nitial temperature Ti " 66° ; 2


19/85

c

289 6.6296.#7<

2.9

h$urve 6.#7<

K

×= =

= =

a,is value is 1.2& curve value is 6.#7


20/85


21/85

+. #hat is meant &! laminar flow and tur&ulent flow

Laminar flow; $aminar flow is sometimes called stream line flow. 0n this type of flow& the fluid

moves in layers and each fluid particle follows a smooth continuous path. The fluid particles in

each layer remain in an orderly se?uence without mi,ing with each other.

Tur&ulent flow; 0n addition to the laminar type of flow& a distinct irregular flow is fre?uency

oserved in nature. This type of flow is called turulent flow. The path of any individual particle isEig – Eag and irregular. 3ig. shows the instantaneous velocity in laminar and turulent flow.


22/85


23/85

.79 2 *76 26+

> Area width length # 2 2@

Q 1 " mm.

1.

Kinematic viscosity v " #


24/85

3or 3lat plate& laminar flow&

1.

6.9

h,

6.9

1

h,

9 , *Ce+

" 9 6.1 *9.6# #6 +

7.< #6 m

δ

δ

−

−

−

= × ×

× × ×= ×

2. Thermal &oundar! la!er thic0ness;

( )

6.111

T h,

1 6.111

T

1

T

*4r+

7.< #6 *6.7=8+


25/85

[ ]

,,

,

1

2

,

2

,

h $(u

K

h 6.179.= , " $ " 6.1m

21.27 #6

h 7.26 W/m K

$ocal heat transfer coefficient h 7.26 W / m K

−

×=

×=

×⇒ =

=

Q

+. Aera%e heat transfer coefficient (h/;

,

2

h 2 h

2 7.26

h #2.# W / m K

= ×

= ×

=

,. cm and 3 > ) cm.

:ien; 3luid temperature T∞ " 16°

Helocity " 2 m/s

$ength $ " 2 m

Wide W W " #.9 m

To find;

1. oundar! la!er thic0ness2. Total dra% force.". Total mass flow rate throu%h the &oundar! la!er &etween 3 > cm and 3 > ) cm.

Solution; 4roperties of air at 16°


26/85

1

7 2

#.#79 %g/m

v #7 #6 m / s

4r 6.


27/85

6.9

f$

9 6.9

-1

f$

#.128 *Ce+

" #.128 *2.9 #6 +

2.79 #6

−

−

=

× ×

= ×

We %now

f$ 2

-1

2

-1 2

-1

t

2

t 2.79 #6

#.#79 *2+

2

Average shear stress t " 7.# #6 (/ mBrag force " Area Average shear stress

" 2 #.9 7.# #6

Brag force " 6.6#8 (

Brag force on two sides of the plate

" 6.6#8

ρ =

⇒ × =×

⇒ ××

× × ×

× 2" 6.617 (

Total mass flow rate etween , " 6 cm and ," 89 cm.

[ ]h, h,9

m 89 68

ρ δ δ ∆ = = − =

!ydrodynamic oundary layer thic%ness

6.9

h, 6.9

6.99 , *Ce+

, " 9 6.89

v

δ −

=−

= × ×× × ×


28/85


29/85

Kinematic viscosityatm

atm

given

4

4ν ν = ×

[ ]

7

1 2

9 27

1 1

# ar 12.= #6

7 #6 ( /m

Atmospheric pressure " # ar

#6 ( /m12.= #6

7 #6 (/m

ν −

−

⇒ = ××

= × ××

Q

9 2

- 2

# ar " # #6 (/m

Kinematic viscosity v " 9.#9 #6 m / s.

× ×

Q

We %now& Ceynolds numer$

Cev

=

9

9

7 #9.#9 #6

Ce #.#6 #6 9 #6

Dince CeJ 9 #6 & flow is laminar

−×=×

= × − ×

×

3or plate& laminar flow&

$ocal nusselt numer

6.9 6.111

,

6.9 6.111

,

( 6.112 *Ce+ *4r+

6.112 *#.#6 #6 + *6.78#+( 16.71

=

= ×=

We %now (, ",h $

K

,

1

h #16.71 > $ " # m@

1


30/85

w

We %now

!eat transferred Q h A *T T +

2.1# *# 6.9+ *971 11+

Q 29.# W

∞= −

= × × × −=

!eat transfer from oth side of the plate " 2 × 29.#

" 968.2 W.

). Air at °C flows oer a flat 'late6 . m lon% at a elocit! of ) m4s. The 'late surface

is maintained at "°C. Determine the heat transferred from the entire 'late len%th to air

ta0in% into consideration &oth laminar and tur&ulent 'ortion of the &oundar! la!er. Also

calculate the 'ercenta%e error if the &oundar! la!er is assumed to &e tur&ulent nature

from the er! leadin% ed%e of the 'late.

:ien ; 3luid temperature T∞ " 6°& $ength $ " 6.8 m& Helocity " 96 m/s & 4late surface

temperature Tw " 166°

To find ;

#. !eat transferred forN

i. :ntire plate is considered as comination of oth laminar and turulent flow.ii. :ntire plate is considered as turulent flow.

2. 4ercentage error.

Solution; We %now 3ilm temperature wf T T

T2

∞−= T

f

1

7 2

1

166 61 K

2

T #


31/85

Case (i/; $aminar – turulent comined. >0t means& flow is laminar upto Ceynolds numer value

is 9 × #69& after that flow is turulent@

Average nusselt numer " (u " *4r+6.111 *Ce+6.8 – 8


32/85

Average heat transfer coefficientO h " ##1.# W/m2K

We %now !eat transfer Q2 " h × A × *Tw ; T∞+

" h × $ × W × *Tw ; T∞+

" ##1.# × 6.8 × # *166 – 6+

Q2 " 2198=.2 W

2 #

#

Q Q2. 4ercentage error "

Q

2198=.2 - #71=6. " #66

#71=6.

" 1.=F

−

×

7. 2) 5%4hr of air are cooled from 1°C to "°C &! flowin% throu%h a ".) cm inner

diameter 'i'e coil &ent in to a heli3 of .+ m diameter. Calculate the alue of air side heat

transfer coefficient if the 'ro'erties of air at +)°C are

5 > .2 #4m5

µ > ." 5%4hr G m

Pr > .,

ρ > 1. 5%4m"

:ien ; ass flow rate in " 269 %g/hr

269Kg/ s in " 6.697 Kg/s

1766=

0nlet temperature of air Tmi " #66°

'utlet temperature of air Tmo " 16°

Biameter B " 1.9 cm " 6.619 m

ean temperature mi momT T

T 79 2

+= = °


33/85

To find; !eat transfer coefficient *h+

Solution;

Ceynolds (umer Ce "B

ν

Kinematic viscosity µ

ν ρ

=

1

< 2

6.661Kg/ s m

1766#.6 Kg/m

v


34/85

We %now that&hB

(uK

=

h 6.619277#.<

6.62=8

×=

!eat transfer coefficient h " 2277.2 W/m2K


35/85

m

1

-7 2

-1

#7 12

2

T 2

4roperties of air at 2 N

" #.7# Kg/m

" #9.= #6 m / s4r " 6.


36/85

1 6.8 6. (u " 6.621 *19.1 #6 + *6.


37/85

Dince Ce J 2166 flow is turulent

$ 26

B 6.696

$#6 66

B

= =

< <

3or turulent flow& *Ce J #6666+

6.699

6.8 6.11

6.699

6.8 6.11

-1

2

B(usselt numer (u " 6.617 *Ce+ *4r+

$

6.696(u 6.617 *9


38/85

wf

f

1

-7 2

-1

f

T T3ilm temperature T

2

767 #67

2

T 197

4roperties of air at 197 " 196

" 6.977 Kg/m

99.7 #6 m / s

4r " 6.7#.68 #6 @

(u "


39/85

-1

h


40/85

wf

f

1

-7 2

1


2

#96


41/85

= 6.111 (u " 6.#9 >9.2= #6 @

(u " 29=.#

+⇒ ×⇒

We %now that&

u c

u

1

2

u

h $(usselt numer (u "

K

h 6.6929=.#

781 #6

h " 191.7 W/m K

−

×=

×

pper surface heated& heat transfer coefficient hu " 191.7 W/m2K

For horiIontal 'late6 lower surface heated;

(usselt numer (u " 6.2< >Mr [email protected]

= 6.29

# c

# c

(u " 6.2< >9.2= #6 @

(u "


42/85

" *191.7 ; ==.7+ × 6.#6 × *#96 –


43/85

#. Birect contact heat e,changers2. 0ndirect contact heat e,changers1. Durface heat e,changers. 4arallel flow heat e,changers9. ounter flow heat e,changers7. ross flow heat e,changers


44/85

We %now that the temperature difference etween the hot and cold fluids in the heat

e,changer varies from point in addition various modes of heat transfer are involved. Therefore

ased on concept of appropriate mean temperature difference& also called logarithmic mean

temperature difference& also called logarithmic mean temperature difference& the total heat

transfer rate in the heat e,changer is e,pressed as

Q " A *∆T+m Where – 'verall heat transfer coefficient W/m2K A – Area m2

*∆T+m – $ogarithmic mean temperature difference.

1+. #hat is meant &! Foulin% factor

We %now the surfaces of a heat e,changers do not remain clean after it has een in use

for some time. The surfaces ecome fouled with scaling or deposits. The effect of these

deposits the value of overall heat transfer coefficient. This effect is ta%en care of y introducing

an additional thermal resistance called the fouling resistance.

1,. #hat is meant &! effectieness

The heat e,changer effectiveness is defined as the ratio of actual heat transfer to the

ma,imum possile heat transfer.

:ffectiveness Actual heat transfer

a,imum possile heat transferε = "

ma,

Q

Q

4art-5

#. #ater is &oiled at the rate of 2 0%4h in a 'olished co''er 'an6 " mm in diameter6 at

atmos'heric 'ressure. Assumin% nucleate &oilin% conditions calculate the tem'eratureof the &ottom surface of the 'an.

:ien ;

m " 2 %g / h

1

2 %g

1766 s

m 7.7 #6 %g/ s−

=

= ×

d " 166 mm " .1m

Solution;


45/85

We %now saturation temperature of water is #66°

i.e. Tsat " #66°

4roperties of water at #66°

3rom !T data oo% 4age (o.#1

1

-7 2

r

-7

Bensity l " =7# %g/m

Kinematic viscosity v " 6.2=1 #6 m / s

4r andtl numer 4 #.


46/85

1

l v

#.<

r

g * +Q pl T!eat flu, l hfg

A sf hfg4

We %now transferred Q " m hfg

!eat transferred Q " m hfg.

ρ ρ µ

σ

× − × ∆= × ×

×

××

Q mhg A A

=

1 1

2

-1 1

2

1 2

Q 7.7 #6 2297.= #6

Ad

7.7 #6 2297.= #6 "

*.1+

Q

2#6 #6 w /m A

surface tension for li?uid vapour interface

π

π

σ

−× × ×=

× × ×

= ×=

At #66° *3rom !T data oo% 4age (o.#


47/85

1

w sat

w

w

2#7 T 6.829


48/85

7

7 2

l" l v =7# 6.2=1 #6

l 28#.9< #6 (s/m

3rom steam Tale at #66

µ ρ

µ

−

−

× = × ×

= ×°

C.D. Khurmi Dteam tale 4age (o.

1

1

g

1

g

hfg 2297.= % /%g

hfg " 2297.= #6 /%g

v #.7


49/85

We %now

!eat transferred Q " H × 0

7

7

-1

Q H 0

A A

H 266#.92 #6 A " d$d$

H 266#.92 #6 "

#.9 #6 .96

H #


50/85

h " 9.97 *∆T+1 3rom !T data oo% 4age (o.#28

h " 9.97 *Tw – Tsat+1

" 9.97 *##9 – #66+1

2h #8


51/85

Case (ii/

4 " 26 arR ∆T " #9°

!eat transfer coefficient h " 9.97 *∆T+1 " 9.97 *#9+1

2h #8


52/85

We %now saturation temperature of water is #66°

i.e. Tsat " #66°

*3rom C.D. Khurmi steam tale 4age (o.

hfg " 2297.=%/%g

hfg " 2297.= × #61 /%g

We %now

w satf

f


2

76 #66

2

T 86

+=

+=

= °

4roperties of saturated water at 86°

*3rom !T data oo% 4age (o.#1+

1

7 2

- =


53/85

6.29

sat w

2

fg

K , *T T +,

g h

µ δ

ρ

× × −= ÷ ÷× ×

Where

" $ " 6.9 m

7 1

, 1 2

,

19.91 #6 778.< #6 6.9 #66 76

=.8# 2297.= #6 =


54/85

. ondensate mass flow rate m

We %now

fg

fg

1

Q m h

Qmh

#.21.287m

2297.= #6

m 6.69 %g/s

= ×

=

=×

=

1. Steam at . &ar is arran%ed to condense oer a ) cm s$uare ertical 'late. The

surface tem'erature is maintained at 2°C. Calculate the followin%.

a. Film thic0ness at a distance of 2) cm from the to' of the 'late.&. Local heat transfer coefficient at a distance of 2) cm from the to' of the 'late.c. Aera%e heat transfer coefficient.d. Total heat transfer e. Total steam condensation rate.

f. #hat would &e the heat transfer coefficient if the 'late is inclined at "°C with

horiIontal 'lane.

:ien ;

4ressure 4 " 6.686 ar

Area A " 96 cm × 96 cm " 96 × 696 " 6.29 m2

Durface temperature Tw " 26°

Bistance , " 29 cm " .29 m

Solution


55/85

4roperties of steam at 6.686 ar

*3rom C.D. Khurmi steam tale 4age no.


56/85

. $ocal heat transfer coefficient h, Assuming $aminar flow

,

1

,

2

%h

,

7#2 #6h #.7 #6

h, =# W/m K

δ −

−

=

×= ×

=

c. Average heat transfer coefficient h

*Assuming laminar flow+

6.291 2

fg

sat w

% g hh 6.=1

$ T T

ρ

µ

× × ×=

× × −

The factor 6.=1 may e replaced y #.#1 for more accurate result as suggested y c adams

6.291 2

fg

sat w

% g hh 6.=1

$ T T

ρ

µ

=

× × −

Where $ " 96 cm " .9 m

6.291 1 2 1

7

2

*7#2 #6 + *==


57/85

sat wh A *T T +

99==.7 6.29 *#.91 26

Q 16.#1=.8 W

× × −

= × × −=

e. Total steam condensation rate *m+

We %now

!eat transfer

fg

fg

Q m h

Qm

h

= ×

=

16.#1=.8m

261.2 #61

m 6.6#29 %g /s

=×

=

f. 0f the plate is inclined at θ with horiEontal

( )

#/

inclined vertical

#/

inclined vertical#/

inclined

2

inclined

h h sin

h h *sin16+

#h 99==.72

h .


58/85

Do our assumption laminar flow is correct.

). A condenser is to desi%ned to condense + 0%4h of dr! saturated steam at a 'ressure

of .12 &ar. A s$uare arra! of tu&es6 each of mm diameter is to &e used. The tu&esurface is maintained at "°C. Calculate the heat transfer coefficient and the len%th of

each tu&e.

:ien ;

766m 766 %g/h " %g /s 6.#77 %g/s

1766

m " 6.#77 %g/s

= =

4ressure 4 – 6.#2 ar (o. of tues " 66

-1

w

Biameter B " 8mm " 8 #6 m

Durface temperature T 16

×= °

Solution

4roperties of steam at 6.#2 ar 3rom C.D. Khurmi steam tale 4age (o.<

sat

fg

1

fg

T =.9

h 218.1 %/%g

h " 218.= #6 /%g

= °

=

×

We %now

w satf

f


2

16 =.9

2

T 1=.


59/85

ρ

ν

µ ρ ν

µ

−

×

= ×

× × ×

×

1

-7 2

1

-7

-7 2

- ==9 %g/m

" .79< #6 m / s

% 728.< #6 W/m%

" " ==9 6.79< #6

" 791.< #6 (s/m

with 66 tues a 26 × 26 tue of s?uare array could e formed

i.e. ( 66 26= =

( 26=

3or horiEontal an% of tues heat transfer coefficient.

6.291 2 fg

sat w

K g hh " 6.


60/85

4rolems on 4arallel flow and ounter flow heat e,changers

3rom !T data oo% 4age (o.#19

3ormulae used

#. !eat transfer Q " A *∆T+m

Where

– 'verall heat transfer coefficient& W/m2K

A – Area& m2

*∆T+m – $ogarithmic ean Temperature Bifference. $TB

3or parallel flow

# # 2 2m

# #

2 2

*T t + *T t +* T+

T t0nT t

− − −∆ =

− −

0n ounter flow

# # 2 2m

# #

2 2

*T t + *T t +* T+

T t0n

T t

− − −∆ =

− −

Where

T# – :ntry temperature of hot fluid ° T2 – :,it temperature of hot fluid °

T# – :ntry temperature of cold fluid ° T2 – :,it temperature of cold fluid °

2. =c

h ph # 2 c pc 2 #m *T T + m *t t +− = −

h – ass flow rate of hot fluid& %g/s

c – ass flow rate of cold fluid %g/s

ph – Dpecific heat of hot fluid G/%g K

pc – Dpecific heat of cold fluid G/%g $

". Surface area of tu&e

A " πB# $


61/85

Where B# 0nner din

. = > m × hf%

Where hfg – :nthalpy of evaporation /%g K

). Bass flow rate

m " ρ A

nit- Cadiation

1. Define emissie 'ower HE and monochromatic emissie 'ower. H&λ

E

The emissive power is defined as the total amount of radiation emitted y a ody per unit

time and unit area. 0t is e,pressed in W/m2.

The energy emitted y the surface at a given length per unit time per unit area in all

directions is %nown as monochromatic emissive power.

2. #hat is meant &! a&sor'tiit!6 reflectiit! and transmissiit!

Asorptivity is defined as the ratio etween radiation asored and incident radiation.

Ceflectivity is defined as the ratio of radiation reflected to the incident radiation.

Transmissivity is defined as the ratio of radiation transmitted to the incident radiation.

". #hat is &lac0 &od! and %ra! &od!5lac% ody is an ideal surface having the following properties.

A lac% ody asors all incident radiation& regardless of wave length anddirection. 3or a prescried temperature and wave length& no surface can emit moreenergy than lac% ody.

0f a ody asors a definite percentage of incident radiation irrespective of their wave

length& the ody is %nown as gray ody. The emissive power of a gray ody is always less than

that of the lac% ody.

. State Planc0’s distri&ution law.The relationship etween the monochromatic emissive power of a lac% ody

and wave length of a radiation at a particular temperature is given y the followinge,pression& y 4lanc%.

9#

2

e #

:

T

λ

λ

λ

−

−

= ÷

Where :λ " onochromatic emissive power W/m2

λ " Wave length – m

c# " 6.1


62/85


63/85


64/85

Solution; 1. Bonochromatic Hmissie Power ;

3rom 4lanc%)s distriution law& we %now

9

#

2

e #

:

T

λ

λ

λ

−

−

= ÷

>3rom !T data oo%& 4age (o.


65/85

" 9.7< × #6-8 W/m2K

⇒ : " *9.7< × #6-8+ *1666+

: " .9= × #67 W/m2

). Total emissie 'ower of a real surface;

*:+real " ε σ T

Where ε " :missivity " 6.89

*:+real "8 6.89 9.7< #6 *1666+−× × ×

7 2

real*: + 1.=6 #6 W /m= ×

2. Assumin% sun to &e &lac0 &od! emittin% radiation at + 5 at a mean distance of

12 × 11 m from the earth. The diameter of the sun is 1.) × 1 m and that of the earth is

1".2 × 1+ m. Calculation the followin%.

#. Total energy emitted y the sun.2. The emission received per m2 ust outside the earth)s atmosphere.1. The total energy received y the earth if no radiation is loc%ed y the earth)s

atmosphere.

. The energy received y a 2 × 2 m solar collector whose normal is inclined at 9° to the

sun. The energy loss through the atmosphere is 96F and the diffuse radiation is 26F of direct radiation.

:ien; Durface temperature T " 7666 K

Bistance etween earth and sun C " #2 × #6#6 m

Biameter on the sun B# " #.9 × #6= m

Biameter of the earth B2 " #1.2 × #67 m

Solution;1. Hner%! emitted &! sun H& > σ T

-8

-8 2

: " 9.7< #6 *7666+

> " Dtefan - 5oltEmann constant

" 9.7< #6 W /m K @

σ

⇒ × ×

×

Q

7 2

2

# #

2=

#8 2

#

: "


66/85


67/85

2

6.96 2899.9

#2


68/85

where

# 2

#2# 2

# # # #2 # 2

T TQ

# ##

A A 3 A

σ

ε ε

ε ε

− =− −

+ +>3rom e?uation (o.*7+@

3or lac% ody # 2 #ε ε = =

#2 # 2 # #2

8 #2

9

#2 #2

Q >T T @ A 3

" 9.7< #6 *#2


69/85


70/85

We %now that&

3## ; 3#2 ; 3#1 " # 5ut& 3## " 6

#1 #2

#1

#1

3 # 3

3 # 6.<

3 6.91

⇒ = −

⇒ = −

=

Dimilarly& 32# ; 322 ; 321 " # We %now 322 " 6

21 2#

21 #2

#1

21

3 # 3

3 # 3

3 " # - 6.<

3 6.91

⇒ = −

⇒ = −

=

3rom electrical networ% diagram&

# #1

2 21

# #2

# #6.1# ....*#+

A 3 7 6.91

# #6.1# ....*2+

A 3 7 6.91

# #6.19 ....*1+

A 3 7 6.<

= =×

= =×

= =×

3rom Dtefan – 5oltEmann law& we %now

[ ]

# #

-8

1 2

#

: T

: T

" 9.7< #6 821

: 27.6# #6 W /m .....*+

σ

σ

=

=

×

= ×

[ ]

[ ]

2 2

-8

1 2

2

1 1

-8

2

1 1

: T

" 9.7< #6 821

: .2 #6 W /m .....*9+

: T

" 9.7< #6 168

: G 9#6.29 W /m .....*7+

σ

σ

=

×= ×

=

×

= =


71/85

>3rom diagram@

The radiosities& G# and G2 can e calculated y using Kirchoff)s law.

⇒ The sum of current entering the node G# is Eero.

At (ode G#N

# # 1 #2 #

# #2 # #1

: G : GG G6

# #6.16=

A 3 A 3

>3rom diagram@

− −−+ + =

1

# 2 # #

1 # 2 # #

1

# 2

27.6# #6 G G G 9#6.29 G 6

6.16= 6.19 6.1#G G G G

8.#< #6 #729 66.16= 6.19 6.19 6.19

-=.2G 2.82G 89.


72/85

!eat lost y plate *#+ is given y

# ##

#

# #

: GQ

#

A

ε

ε

−=

− ÷

1 1

#

1

#

27.6# #6 #6.


73/85


74/85


75/85


76/85

2

2 2

2

2 2

2 2

2 2

! o

! o '

! o

! o '

' m ! ' m

' m ! ' m

4 6.##.69

4 4 6.# 6.2

46.111

4 4

4 $ 4 $ 6.67 6.614 $ 4 $ 6.6=

= =+ +

=+

× + × = +× + × =

3rom !T data oo%& 4age (o.=9& we can find correction factor for mi,ture of '2 and 2! o.

nit-9 ass Transfer

1. #hat is mass transfer

The process of transfer of mass as a result of the species concentration difference in ami,ture is %nown as mass transfer.

2. :ie the e3am'les of mass transfer.

Dome e,amples of mass transfer.

#. !umidification of air in cooling tower 2. :vaporation of petrol in the caruretor of an 0 engine.1. The transfer of water vapour into dry air.

". #hat are the modes of mass transfer

There are asically two modes of mass transfer&

#. Biffusion mass transfer 2. onvective mass transfer

. #hat is molecular diffusion

The transport of water on a microscopic level as a result of diffusion from a region of

higher concentration to a region of lower concentration in a mi,ture of li?uids or gases is %nown

as molecular diffusion.

). #hat is Hdd! diffusion

When one of the diffusion fluids is in turulent motion& eddy diffusion ta%es place.

+. #hat is conectie mass transfer

onvective mass transfer is a process of mass transfer that will occur etween surface

and a fluid medium when they are at different concentration.


77/85

,. State Fic0’s law of diffusion.

The diffusion rate is given y the 3ic%)s law& which states that molar flu, of an element

per unit area is directly proportional to concentration gradient.

a aa

2

2

a

1a

m dB A d,

where&

ma %g -moleolar flu,&

A s-m

B Biffusion coefficient of species a and & m / s

dconcentration gradient& %g/m

d,

= −

−

−

. #hat is free conectie mass transfer

0f the fluid motion is produced due to change in density resulting from concentration

gradients& the mode of mass transfer is said to e free or natural convective mass transfer.

:,ample N :vaporation of alcohol.

. Define forced conectie mass transfer.

0f the fluid motion is artificially created y means of an e,ternal force li%e a lower or fan&

that type of mass transfer is %nown as convective mass transfer.

:,ampleN The evaluation if water from an ocean when air lows over it.

1. Define Schmidt *um&er.

0t is defined as the ratio of the molecular diffusivity of momentum to the molecular

diffusivity of mass.

olecular diffusivity of momentumDc

olecular diffusivity of mass=

11. Define Scherwood *um&er.

0t is defined as the ratio of concentration gradients at the oundary.

m

a

2

a

h ,Dc

B

hm ass transfer coefficient& m/s

B Biffusion coefficient& m /s

, $ength& m

=

−

−

−


78/85

4art-5

#.

m2 4s. The solu&ilit! of h!dro%en in the mem&rane is 2.1 × 1-" 1

%g mole

m ar

−

An uniform tem'erature condition of 2° is assumed.

alculate the following

#. olar concentration of hydrogen on oth sides2. olar flu, of hydrogen1. ass flu, of hydrogen

:ien Data;

0nside pressure 4# " 1 ar

'utside pressure 42 " # ar

Thic%ness& $ " 6.29 mm " 6.29 × #6-1 m

Biffusion coefficient Ba "8 2=.# #6 m / s−×

-1

1

%g moleDoluility of hydrogen " 2.# #6

m ar

Temperature T " 26

−×

−°

To find

#. olar concentration on oth sides a# and a22. olar flu,

1. ass flu,Solution ;

#. olar concentration on inner side&

a# " Doluility × inner pressure

a2 " 2.# × #6-1 × 1

a# " 7.1 × #6-1 1%g - molem

olar concentration on outer side

a# " soluility × 'uter pressure

a2 " 2.# × #6-1 × #


79/85

a2 " 2.# × #6-1

1

%g - mole

m

2. We %now [ ]o a a# a2m B

A $

= −

[ ]1 1

1

7a

2

=.# *7.1 #6 2.# #6 +olar flu,& " #.2 6.29 #6

m %g-mole #.92 #6

A s-m

− −

−

−

× − ×− −×

= ×

1. ass flu, " olar flu, × olecular weight

7

2

2

-7

2

%g mole#.92 #6 2 mole

s m

> olecular weight of ! is 2@

%gass flu, " 1.6 #6 .s m

− −= × ×−

×−

Q

2. 83!%en at 2)°C and 'ressure of 2 &ar is flowin% throu%h a ru&&er 'i'e of inside

diameter 2) mm and wall thic0ness 2.) mm. The diffusiit! of 82 throu%h ru&&er is .21 ×

1- m2 4s and the solu&ilit! of 82 in ru&&er is ".12 × 1-" 1

%g mole

m ar

−−

. Find the loss of 82 &!

diffusion 'er metre len%th of 'i'e.

:ien data;

Temperature& T " 29° fig

0nside pressure 4# " 2 ar

0nner diameter d# " 29 mm

0nner radius r # " #2.9 mm " 6.6#29 m

'uter radius r 2 " inner radius ; Thic%ness

" 6.6#29 ; 6.6629

r 2 " 6.6#9 m

= 2

a

-1

1

Biffusion coefficient& B 6.2# #6 m / s

%g moleDoluility& " 1.#2 #6

m

−= ×

−×

olar concentration on outer side&

a2 " Doluility × 'uter pressure


80/85

a2 " 1.#2 × #6-1 × 6

a2 " 6

>Assuming the partial pressure of '2 on the outer surface of the tue is Eero@

We %now&

[ ]a a# a2a

2 #2 #

2

#

B m

A $

2 $ *r r +3or cylinders& $ " r r R A "

r 0n

r

π

−=

−−

[ ]

[ ]

a a# a2a

2 # 2 #

a a# a2a

2

#

-##

a

B molar flu,& *#+

2 $*r r + *r r +

2 $.B m > $ength " #m+r

0nr

%g mole m .9# #6 .

s

π

π

−⇒ =

− −

−⇒ =

−= ×

Q

1. An o'en 'an 21 mm in diameter and ,) mm dee' contains water at 2)°C and is

e3'osed to dr! atmos'heric air. Calculate the diffusion coefficient of water in air. Ta0e

the rate of diffusion of water a'our is .)2 × 1- 0%4h.

:ien ;

Biameter d " 2#6 " .2#6 m

Beep *,2 – ,#+ "


81/85


82/85

:ien ;

Biameter& d " #96mm " .#96m

Beep *,2 –,#+ " water ; air@ at 29°

1 2=1 #6 m /h−= ×

12

a

9 2

a

=1 #6 B m / s

1766

B 2.98 #6 m / s .

−

−

×⇒ =

= ×

Atmospheric air 96F C! *2+

We %now that& for isothermal evaporation&

olar flu,&

( )a a w2

2 # w#

2 2

2

9 2

w#

m B 4 440n ......*#+

A MT , , 4 4

where&

A - Area " d *.#96+

Area 6.6#


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4w# " 6.61#77 × #69 (/m2

4w2 " 4artial pressure at the top of the test pan corresponding to 29 ° and 96F relative

humidity.

At 29°

9w2

9

w2

2

w2

4 6.61#77 ar " 6.61#77 #6 6.96

4 6.61#77 #6 6.96

4 #981 (/m

= × ×= × ×

=

a*#+

6.6#3rom !T data oo%& 4age (o.#


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Where& hm – ass transfer coefficient – m/s

3or Turulent flow N

Dhedwood (umer *Dh+ " >.61< *Ce+6.8 – 83rom !T data oo%& 4age (o.#86@

B=

Soled Pro&lems on Flat Plate.

). Air at 1°C with a elocit! of " m4s flows oer a flat 'late. The 'late is ." m lon%.

Calculate the mass transfer coefficient.

:ien ;

3luid temperature& T∞ " #6°

Helocity& " 1 m/s

$ength& , " 6.1 m

To find; ass transfer coefficient *hm+

Solution; 4roperties of air at #6° >3rom !T data oo%& 4age (o.22@

Kinematic viscosity. H " #.#7 × #6-7 m2/s

We %now that&

-7

9 9

9

,Ceynolds (umer& Ce "

1 6.1 "

#.#7 #6

Ce " 6.71 #6 9 #6

Dince& Ce J 9 #6 & flow is laminar

ν

××

× < ×

×

For Laminar flow6 flat 'late6

Dherwood (umer *Dh+ " 6.77 *Ce+6.9 *Dc+6.111 I.*#+

>3rom !T data oo%& 4age (o.#


85/85

Ba – Biffusion coefficient *water;Air+ at #6° " 8°

21

9 2

a

m

Answer 2 & 16 Marks HMT.doc

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